Question

find the point of intersection of two straight lines.A) 3x-y-5=0, x+2y-4=0​

Answer 1

5x−6y−1=0

=>x=

5

6y+1

3x+2y+5=0

=>3(

5

6y+1

)+2y+5=0

=>18y+3+10y+25=0

=>y=−1

x=

5

6y+1

=

5

6×(−1)+1

=−1

Point of intersection (−1,−1)

Required line perpendicular to the line 3x−5y+11=0 is

5x+3y+c=0 where c is a constant

5(−1)+3(−1)+c=0

=>−5−3+c=0

=>c=8

Required line is 5x+3y+8=0

I hope it will help you